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In multivariate statistics, if is a vector of random variables, and is an -dimensional symmetric matrix, then the scalar quantity is known as a quadratic form in .
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It can be shown that[1]
where and are the expected value and variance-covariance matrix of , respectively, and tr denotes the trace of a matrix. This result only depends on the existence of and ; in particular, normality of is not required.
A book treatment of the topic of quadratic forms in random variables is that of Mathai and Provost.[2]
Since the quadratic form is a scalar quantity, .
Next, by the cyclic property of the trace operator,
Since the trace operator is a linear combination of the components of the matrix, it therefore follows from the linearity of the expectation operator that
A standard property of variances then tells us that this is
Applying the cyclic property of the trace operator again, we get
In general, the variance of a quadratic form depends greatly on the distribution of . However, if does follow a multivariate normal distribution, the variance of the quadratic form becomes particularly tractable. Assume for the moment that is a symmetric matrix. Then,
In fact, this can be generalized to find the covariance between two quadratic forms on the same (once again, and must both be symmetric):
In addition, a quadratic form such as this follows a generalized chi-squared distribution.
The case for general can be derived by noting that
so
is a quadratic form in the symmetric matrix , so the mean and variance expressions are the same, provided is replaced by therein.
In the setting where one has a set of observations and an operator matrix , then the residual sum of squares can be written as a quadratic form in :
For procedures where the matrix is symmetric and idempotent, and the errors are Gaussian with covariance matrix , has a chi-squared distribution with degrees of freedom and noncentrality parameter , where
may be found by matching the first two central moments of a noncentral chi-squared random variable to the expressions given in the first two sections. If estimates with no bias, then the noncentrality is zero and follows a central chi-squared distribution.
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